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3 years ago

7

1.) Rn-222 decays from 400 grams to 6.25 grams in 240 minutes. How long is one “half-life.

1 answer:

KiRa [710]3 years ago

7 0

Answer:

40

Explanation:

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An inclined plane is 5.00 m long and 3.00 m high. What is the ideal mechanical advantage of this machine?

Dvinal [7]

<span>inclined plane formula is length/hight

so 5/2= 2.5

and work= f x d so
work= 5 x 104 x 10= 5200 W
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Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys

max2010maxim [7]

Answer:

The spring was compressed the following amount:

$\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount $\Delta x$ ), and the total final energy is the addition of the kinetic energies of both masses:

$E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2$

$E_i=E_f\\$

$\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

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3 years ago

What to forces effects the momentum of an object

faltersainse [42]

Any force can affect the momentum of an object
if it changes the object's speed.

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2 years ago

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If it takes you 30 seconds to move a chair 20 meters across the floor, using a force of 5 Newton’s, how much power did you put o

natta225 [31]

Answer:

3.33 Joules Per Second

Explanation:

Before finding the Power, we need to calculate the Work Done. The Work Done can be calculated using the formula:

WD = F × d

where F is the Magnitude of Force in <em>N</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>d is the Parallel Distance moved by the object in <em>m</em><em>.</em>

Hence, by Applying this formula, we get:

WD = (5)(20)

= 100 J

From here calculating Power is simple as it is the Rate of Work Done. Hence,

Power = 100/30

= <u>3</u><u>.</u><u>3</u><u>3</u><u> </u><u>J</u><u>/</u><u>s</u>

Therefore, the power put out is <u>3</u><u>.</u><u>3</u><u>3</u><u> </u><u>J</u><u>o</u><u>u</u><u>l</u><u>e</u><u>s</u><u> </u><u>p</u><u>e</u><u>r</u><u> </u><u>S</u><u>e</u><u>c</u><u>o</u><u>n</u><u>d</u><u>.</u>

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